WEBVTT
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Let's take a look at the sum and difference formulas. Here I've listed on the board the sum and difference
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formulas for cosine, sine, and tangent. Let's see how we can use these sum and difference formulas to
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find exact values. In this example, I want to find the exact value of the sine of 7 pi over 12, and I'm going
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to use a sum formula. So the problem is 7 pi over 12 of course is not one of your special angles. We want
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to rewrite 7 pi over 12 as the sum of two angles, or the difference of two angles, that are special. So I'll
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write this as the sine of 4 pi over 12 plus 3 pi over 12. Now of course that adds up to 7 pi over 12, and what's
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very important about that is that when you reduce these you get pi over 3, which of course is a special
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angle, and pi over 4, which of course is another special angle, so we can find the exact value. You couldn't
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rewrite this, say, as 5 pi over 12 plus 2 pi over 12 because 5 pi over 12 is not a special angle. So it's very
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important that when you rewrite it these two angles are special. So now we have the sine of an angle plus
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an angle. Well we can use this identity right here. So in this case, my alpha is pi over 3, and my beta is
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pi over 4. So using this identity, I can rewrite this as the sine of pi over 3 cosine pi over 4 plus the cosine
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of pi over 3 times the sine of pi over 4. Don't skip the step where you write the identity. Now we've got
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it down to trig functions of special angles, and you should know these values very well at this point.
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Let's plug in those values. The sine of pi over 3, remember, is square root of 3 over 2. The cosine of pi over 4,
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recall, is 1 over the square root of 2. The cosine of pi over 3 is 1/2, and finally, the sine of pi over 4 is
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1 over the square root of 2. So let's see what we get. We get the square root of 3 over 2 square roots of 2
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plus 1 over 2 square roots of 2. Since we have a common denominator, we can add and get the square
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root of 3 plus 1 over 2 square roots of 2. So the sine of 7 pi over 12, the exact value,
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is square root of 3 plus 1 over 2 square roots of 2. Let's take a look at another example.
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In this example, we want to find the exact value of the tangent of 17 pi over 12 without the use of a
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calculator. This time we're going to think of two numbers that either add or subtract to get 17 pi over 12.
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Well, let's see. 17 pi over 12 could be written as 8 pi over 12 plus 9 pi over 12. That certainly adds
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up to 17, and if I reduce each of these, let's see. That'll be 2 pi over 3 and 3 pi over 4. Those are both
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special. But you could also write it, there are lots of other ways, you could write it as 21 pi over 12 minus
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4 pi over 12. If I reduce each of these, let's see what we get. Let's see, 3, that would be 7 pi over 4
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minus pi over 3. So the way you rewrite it is not unique. Oftentimes there's more than one way. Let's go
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ahead and use this way. So we're going to rewrite this as the tangent of 21 pi over 12 minus 4 pi over 12,
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and that's 7 pi over 4 minus pi over 3. Ok, so now we've got it written as the tangent of alpha minus beta,
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so we can use this identity right here. My alpha is 7 pi over 4. My beta is pi over 3. So that's going to give
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us tangent of 7 pi over 4 minus tangent of pi over 3 all over 1 plus tangent of 7 pi over 4 times the
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tangent of pi over 3. Ok, now we just need to find those values. Again, these should be values that you
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know very well at this point. The tangent of 7 pi over 4, well the reference angle is pi over 4. Tangent of
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pi over 4 is 1, but remember 7 pi over 4 is in quadrant four. The tangent is negative, so this will be
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negative 1. The tangent of pi over 3 is the square root of 3. Plugging in those values, we get negative 1
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minus the square root of 3 over 1 minus the square root of 3. So this is equal to the tangent of 17 pi over 12.
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In this example, we want to find the exact value of this expression without the use of a calculator.
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So we look at it. We notice these angles are not special. But look at our identities over here. Which one
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does this look like? So we see cosine cosine plus sine sine. Well that looks like this one, doesn't it? So in
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this case, here's my alpha, and here's my beta. So this could be rewritten as the cosine of alpha minus beta.
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So we're going to have 7 pi over 12 minus 5 pi over 12. Well that's the same thing as the cosine, of course,
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of 2 pi over 12, which reduces to the cosine of pi over 6, and of course that's much easier to find.
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Remember the cosine of pi over 6 is simply the square root of 3 over 2. So I've taken this complicated
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expression, recognized it as one of these identities, and rewritten it.
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In this example, we're given that the sine of alpha is 3/5. Alpha is between pi over 2 and pi. The tangent
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of beta is negative square root of 5. Beta is between negative pi over 2 and 0. We want to find the exact
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value of the sine of alpha minus beta and the tangent of alpha plus beta. Now what's important here is
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that we draw the angles in the correct quadrant, so notice that alpha is between pi over 2 and pi, so that
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means that alpha is in quadrant two. So we're going to draw a picture for alpha in quadrant two,
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and we know the sine is 3/5. We know that the sine is y over r, so we know this side is 3. This side is 5.
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We use the Pythagorean theorem. The third side is 4, but we know that since we're in the second quadrant
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the x-coordinate is negative, so the x-value is negative 4. The y is 3, and r is 5. Now we'll do the same
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thing for beta. Beta is between negative pi over 2 and 0. Well negative pi over 2 is here, 0 is here, so we
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know beta is in quadrant four. We're going to draw a picture for beta. The tangent is negative square
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root of 5, so we know the y over x is negative square root of 5. So we know we're going to put square
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root of 5 on the y, and we're going to put 1 on the x, and again we know that in the fourth quadrant the
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x is positive, the y is negative, and of course the hypotenuse is always positive, r is always positive.
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Using the Pythagorean theorem, we get this side to be the square root of 6. And now we've got the
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pictures drawn. We've got all of the sides, so now we're ready to begin writing the identitites. So let's do
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part a. We want to find the sine of alpha minus beta. We have to remember that identity goes sine alpha
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cosine beta minus cosine alpha sine beta, and once we have these values, we can find these values
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pretty easily. The sine of alpha, we make sure that we go to alpha, so it's important that you label them.
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The sine of alpha is 3/5. Cosine of beta we go here, x over r, so it's 1 over the square root of 6.
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Cosine of alpha would be negative 4/5. Remember x is negative here. And the sine of beta is y over r,
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so the sine of beta would be negative square root of 5 over the square root of 6. Now that we've got our
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values we can simply combine. What do we get? We get 3 over 5 square roots of 6 minus 4 square roots
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of 5 over 5 square roots of 6. Since we have a common denominator, we can combine the numerators.
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So our final answer is 3 minus 4 square roots of 5 over 5 square roots of 6. Now let's look at part b.
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Now in part b, we want to find the tangent of alpha plus beta, so again we write the identity for that,
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which is tangent of alpha plus tangent of beta over 1 minus tangent alpha tangent beta. And now that
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we've got our pictures still, we'll just use those to find these values. What's the tangent of alpha?
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Remember tangent of alpha is y over x, so we're going to get 3 over negative 4, and the tangent of beta
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would be y over x which is negative square root of 5. Now we can't leave our answer like that because
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we can't leave complex fractions, so we have to simplify that. Really the easiest way to do that is
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to multiply, in this case numerator and denominator by 4. So let's see what we get. We're going to get
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negative 3, and then we're going to get minus 4 square roots of 5. In the denominator, we're going to
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get 4 and two negatives, so we have minus 3 square roots of 5. We can leave our answer like that. We
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can leave radicals in the denominator. We just can't leave complex fractions, so here's our final answer.
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Now let's look at one final example.
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In this example, we want to find the exact value of the cosine of the inverse tangent of 4/3 plus the
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inverse cosine of negative 9 over 41. We can let the inverse tangent of 4/3, we can call that alpha, and
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we'll call the inverse cosine of negative 9 over 41 beta. Here's where it's very important that you know
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the range of these inverse functions. Remember the inverse tangent lies between negative pi over 2 and
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pi over 2, and the inverse cosine lies between 0 and pi. So you can see it is very important that you know
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that and not have to look it up. So now what we have is the cosine of alpha plus beta. Remember that
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identity. We need to again draw pictures for both alpha and beta. Now alpha is the inverse tangent of 4/3.
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Since this number is positive and alpha lies between negative pi over 2 and pi over 2, we know alpha
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is in quadrant one. So we'll draw a picture for alpha. The tangent is 4/3, so the y is 4, the x is 3, and
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again we can use the Pythagorean theorem to find the third side which will be 5, and we'll do the same
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thing for beta. Beta lies between 0 and pi. This number is negative, so we know it lies in quadrant two.
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Cosine is x over r, so this side is 9. This side is 41, and again we use the Pythagorean theorem to find
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the third side. It turns out to be 40. So now that we have our pictures, we'll write the identity for the
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cosine of alpha plus beta, which is cosine alpha cosine beta minus sine alpha sine beta, and then we
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can just plug in our values. What's the cosine of alpha? It will be 3/5. The cosine of beta will be, let's see.
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Cosine of beta will be negative 9 over 41. Sine of alpha would be 4/5, and sine of beta would be 40 over 41.
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So now that we've plugged in our values, we simply determine these values. So that will give you
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negative 27 over 205 minus 160 over 205, which gives you negative 187 over 205 for our final answer.
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Now in these examples, they illustrate how important it is to know the exact values of the trig functions
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of the special angles as well as the range of the inverse functions. So be sure you know that.
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Then practice these examples until you've mastered them.